hdu---(Tell me the area)（几何/三角形面积以及圆面积的一些知识）

Tell me the area

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1876    Accepted Submission(s): 567

Problem Description

    There are two circles in the plane (shown in the below picture), there is a common area between the two circles. The problem is easy that you just tell me the common area.

Input

There are many cases. In each case, there are two lines. Each line has three numbers: the coordinates (X and Y) of the centre of a circle, and the radius of the circle.

Output

For each case, you just print the common area which is rounded to three digits after the decimal point. For more details, just look at the sample.

Sample Input

0 0 2 2 2 1

Sample Output

0.108

Author

wangye

Source

2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（4）

对于平面内，任意两个圆，存在这些关系： 内含和内切，以及相交，外切和外离。

（1）对于内切，我们只需要求出面积最小圆的面积，

（2）对于外切及外离，得到的面积必然为0.0；

 （3）对于相交，那么我们需要求出这些

          ： 知道两点坐标： 求出dist两点之间的距离;

              知道三边，可以求出三边对应的角度： a^2+b^2-2*a*b*cos(g)=dist^2;

        对于四边形的面积： sm=s3(三角形的面积)*2；

                                  s3=sqrt(p*(p-r1)*(p-r2)*(p-d));

    然后求出对应两个扇形的面积：s1，s2  依据： s=1/2*g*r*r;

     最后：  s=s1+s2-sm;

    代码：

 1 #include<cstdio>
 2 #include<cmath>
 3 #include<algorithm>
 4 #include<iostream>
 5 using namespace std;
 6 
 7 struct circle
 8 {
 9   double x,y,r;
10 };
11 double dist(circle a,circle b)
12 {
13   return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
14 }
15 int main()
16 {
17    circle a,b;
18    double d,p,area,sb,sa;
19   while(scanf("%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&a.r,&b.x,&b.y,&b.r)!=EOF)
20   {
21     d=dist(a,b);
22     double rr=min(a.r,b.r);
23     if(d<=abs(a.r-b.r))   //内含或者内切
24        area=acos(-1.0)*rr*rr;
25     else
26        if(d>=a.r+b.r)
27           area=0.0;
28     else{
29          p=(a.r+b.r+d)/2.0;
30       sa=acos((a.r*a.r+d*d-b.r*b.r)/(2.0*a.r*d));
31       sb=acos((b.r*b.r+d*d-a.r*a.r)/(2.0*b.r*d));
32         area=sa*a.r*a.r+sb*b.r*b.r-2*sqrt(p*(p-a.r)*(p-b.r)*(p-d));
33     }
34     printf("%.3lf\n",area);
35   }
36  return 0;
37 }